$\int\frac{\cos 4x-1}{\cot x-\tan x}dx$ is equal to:

$\frac{-1}{2}\cos(4x)+C$ $\frac{-1}{4}\cos(4x)+C$ $\frac{-1}{2}\sin(2x)+C$ None of these

None of these

$I=\int\frac{\cos 4x-1}{\cot x-\tan x}dx=\int(\frac{-2\sin^2(2x)}{\cos^2x-\sin^2x}).\sin x.\cos x\,dx$ $=-\int\frac{1-\cos^2(2x)}{\cos (2x)}.\sin (2x)dx=\frac{1}{2}\int\frac{1-t^2}{t}.dt=\frac{1}{2}(\log|t|-\frac{t^2}{2})+C=\frac{1}{2}\log|\cos 2x|-\frac{\cos^2(2x)}{4}+C$